Control Systems

6% of exam

Block diagrams, Bode plots, closed-loop stability, steady-state error, transient performance.

4 concepts

Exam areas

A. Block diagrams

Block Diagrams and Signal Flow

Reduce series, parallel, and feedback blocks to one transfer function, then read off closed-loop gain, sensitivity, and disturbance rejection.

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B. Bode plots

Bode Plots

Build straight-line magnitude and phase asymptotes from poles and zeros, then read gain margin, phase margin, and stability straight off the plot.

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C. Closed-loop response, open-loop response, and stability

Closed-Loop Response and Stability

Find pole locations from the characteristic equation, test stability with Routh-Hurwitz and root locus, and tie BIBO stability to margins.

Stability is the first question any control system must answer: will a bounded input produce a bounded output, or will the response blow up? Everything reduces to where the closed-loop poles sit in the complex

s

-plane. The FE handbook states the rule that anchors this whole topic — the closed-loop characteristic equation is

1+G_1(s)G_2(s)H(s)=0

, and its roots are the poles whose locations decide stability. This concept is the storefront for the controls unit because nearly every C-area question is a disguised pole-location problem.

The characteristic equation

For the loop

1+G(s)H(s)=0

, clearing fractions gives a polynomial in

s

whose roots are the closed-loop poles. Open-loop poles are the roots of the forward/loop transfer function's denominator; closing the loop moves them. Set the denominator of

T(s)=G/(1+GH)

to zero — that is the characteristic equation, and its roots, not the open-loop poles, govern the closed-loop response.

1 + G(s)H(s) = 0 \;\Longleftrightarrow\; \text{denominator of } T(s) = 0

Pole locations and BIBO stability

A continuous LTI system is bounded-input bounded-output (BIBO) stable if and only if every closed-loop pole has a strictly negative real part — all poles in the open left half-plane. A pole on the imaginary axis (

\text{Re}=0

) gives a marginally stable, sustained oscillation; any pole in the right half-plane gives an unbounded, growing response. Complex poles

-\sigma\pm j\omega_d

produce a decaying (

\sigma>0

) or growing (

\sigma<0

) sinusoid; the real part sets the envelope and the imaginary part sets the ringing frequency.

\text{BIBO stable} \iff \text{Re}(p_i) < 0 \ \text{for all poles } p_i

Routh-Hurwitz criterion

Routh-Hurwitz tells you how many roots lie in the right half-plane without factoring the polynomial. Build the Routh array from the characteristic polynomial

a_n s^n + \cdots + a_0

: the first two rows hold alternating coefficients, and each lower entry is the negative of the

2\times 2

determinant of the two rows above, divided by the pivot (equivalently, cross-multiply in the order shown in the equation below). The system is stable if and only if every entry in the first column is present and has the same sign; the number of sign changes equals the number of right-half-plane roots. A necessary first check: all coefficients must be present and positive.

b_i = \dfrac{a_{n-1}a_{n-2i}-a_n a_{n-2i-1}}{a_{n-1}}

Finding the marginal gain

The most common Routh question gives a gain

K

inside the characteristic polynomial and asks for the range of

K

that keeps the system stable, or the value of

K

at the edge of instability. Force a first-column entry to zero to find the marginal gain; the auxiliary equation formed from the row above that zero then gives the frequency of the sustained oscillation. This is the algebraic twin of a pole crossing the imaginary axis.

\text{first-column entry} = 0 \;\Rightarrow\; K_{\text{marg}}, \quad A(s)=0 \;\Rightarrow\; \omega_{\text{osc}}

Root locus

The root locus traces how the closed-loop poles migrate in the

s

-plane as a gain

K

sweeps from

0

\infty

. The locus starts at the open-loop poles (

K=0

) and ends at the open-loop zeros or at infinity (

K\to\infty

). A branch lies on the real axis to the left of an odd count of real poles and zeros. The system goes unstable at the gain where a branch crosses the imaginary axis — exactly the marginal

K

Routh finds. Root locus is the picture; Routh is the calculation; they agree at the crossing.

1 + K\,\dfrac{N(s)}{D(s)} = 0, \qquad K: 0 \to \infty

Gain margin and relative stability

Being merely stable is not enough; you want margin. The linear gain margin is the factor between the current gain and the marginal gain found above,

\text{GM}_{\text{lin}}=K_{\text{marg}}/K_{\text{design}}

— a bare, unitless ratio. The handbook's dB gain margin is

\text{GM}_{\text{dB}}=-20\log_{10}|G(j\omega_{180})|=20\log_{10}(K_{\text{marg}}/K_{\text{design}})

, read off the Bode magnitude at the phase-crossover frequency

\omega_{180}

. Keep the two forms distinct: a linear factor of

5

corresponds to

20\log_{10}5\approx 14\,\text{dB}

, not

5\,\text{dB}

, so never set the bare ratio equal to the dB value. A large gain margin means the poles sit well inside the left half-plane; a small one means a modest gain increase (or modeling error) could push them across. Relative stability is quantified by both gain and phase margins.

\text{GM}_{\text{lin}} = \dfrac{K_{\text{marg}}}{K_{\text{design}}}, \qquad \text{GM}_{\text{dB}} = 20\log_{10}\!\dfrac{K_{\text{marg}}}{K_{\text{design}}} = -20\log_{10}\big|G(j\omega_{180})\big|\ \text{(dB)}

Exam strategy

First form the characteristic equation by setting

1+GH=0

and clearing fractions — never test the open-loop denominator. For a numeric polynomial, scan for missing or negative coefficients (instant instability) before building the Routh array. When

K

appears, write the array symbolically, set the first-column term to zero for

K_{\text{marg}}

, and use the auxiliary equation for the oscillation frequency. Reserve root-locus reasoning for 'which way do the poles move' questions. Keep BIBO crisp: strictly-left-half-plane poles only — a pole on the

j\omega

axis is marginal, not stable.

Key equations

Closed-loop characteristic equation

1 + G_1(s)G_2(s)H(s) = 0

Roots are the closed-loop poles. For unity feedback, $1+G(s)=0$ . Set the denominator of $T(s)$ to zero.

Closed-loop transfer function

T(s) = \dfrac{G(s)}{1+G(s)H(s)}

Poles = roots of $1+GH$ ; these (not the open-loop poles) determine the closed-loop response.

BIBO stability condition

\text{Re}(p_i) < 0 \quad \forall\, i

All closed-loop poles strictly in the left half-plane. A $j\omega$ -axis pole is marginal; any RHP pole is unstable.

Routh array entry

b_i = \dfrac{a_{n-1}a_{n-2i} - a_n a_{n-2i-1}}{a_{n-1}}

Each entry is the negative of the $2\times 2$ determinant of the two rows above, divided by the pivot $a_{n-1}$ — i.e. cross-multiply in the order shown in the latex (pivot-row column term minus top-row column term). Continue until the $s^0$ row.

Routh stability test

\text{stable} \iff \text{first column all same sign}

Number of first-column sign changes = number of right-half-plane roots.

Necessary coefficient condition

a_n, a_{n-1}, \dots, a_0 > 0 \ \text{(all present)}

Necessary (not sufficient): any missing or non-positive coefficient means at least one non-LHP root.

Auxiliary equation (marginal case)

A(s) = c\,s^2 + d = 0 \;\Rightarrow\; \omega_{\text{osc}} = \sqrt{d/c}

Formed from the row above a zeroed first-column entry; gives the sustained-oscillation frequency at $K_{\text{marg}}$ .

Root-locus condition

1 + K\dfrac{N(s)}{D(s)} = 0

Loci start at open-loop poles ( $K=0$ ), end at open-loop zeros or infinity ( $K\to\infty$ ).

Gain margin (dB)

\text{GM}_{\text{dB}} = -20\log_{10}\big|G(j\omega_{180})\big| = 20\log_{10}\dfrac{K_{\text{marg}}}{K_{\text{design}}}

Handbook dB form, evaluated at the phase-crossover $\omega_{180}$ . The linear gain margin is the bare factor $K_{\text{marg}}/K_{\text{design}}$ ; the dB value is $20\log_{10}$ of that ratio (e.g. a factor of $5\approx 14$ dB). Positive dB = stable, minimum-phase.

Worked examples

Stable gain range by Routh-Hurwitz

Problem. A unity-feedback system has open-loop transfer function $G(s)=\dfrac{K}{s^3+2s^2+s}$ , giving the characteristic equation $1+G(s)=0\Rightarrow s^3+2s^2+s+K=0$ . Find the range of $K$ for stability and the oscillation frequency at the stability limit.

Solution. Routh array for

s^3+2s^2+s+K

s^3:\ 1,\ 1

s^2:\ 2,\ K

s^1:\ \dfrac{(2)(1)-(1)(K)}{2}=\dfrac{2-K}{2},\ 0

s^0:\ K

First column all positive requires

\dfrac{2-K}{2}>0

and

K>0

, so

0<K<2

. Marginal at

K=2

: auxiliary equation from the

s^2

row is

2s^2+K=2s^2+2=0\Rightarrow s^2=-1\Rightarrow \omega_{\text{osc}}=1\,\text{rad/s}

. Sanity check: at

K=2

a pole pair sits on the

j\omega

axis at

\pm j1

, the boundary between stable and unstable — consistent with sustained

1\,\text{rad/s}

oscillation.

s^1\ \text{row}: \dfrac{2-K}{2} > 0 \;\Rightarrow\; 0 < K < 2

Count right-half-plane roots

Problem. Is the system with characteristic polynomial $s^4+2s^3+3s^2+4s+5=0$ stable? If not, how many unstable poles?

Solution. Routh array:

s^4:\ 1,\ 3,\ 5

s^3:\ 2,\ 4

s^2:\ \dfrac{(2)(3)-(1)(4)}{2}=1,\ 5

s^1:\ \dfrac{(1)(4)-(2)(5)}{1}=-6

s^0:\ 5

First column:

1,\ 2,\ 1,\ -6,\ 5

. Sign goes

+\to+\to+\to-\to+

: two sign changes. Two right-half-plane roots, so the system is unstable. Sanity check: all coefficients are positive, so the necessary condition passes — but Routh exposes the two RHP roots that the coefficient test alone would miss.

\text{first column: } 1,\,2,\,1,\,-6,\,5 \;\Rightarrow\; 2\ \text{RHP roots}

Closed-loop poles from the characteristic equation

Problem. A unity-feedback loop has $G(s)=\dfrac{8}{s+2}$ . Find the closed-loop pole and state whether the system is BIBO stable.

Solution. Characteristic equation:

1+G(s)=0\Rightarrow 1+\dfrac{8}{s+2}=0\Rightarrow (s+2)+8=0\Rightarrow s=-10

. The single closed-loop pole is at

s=-10

, strictly in the left half-plane. BIBO stable. Sanity check: the open-loop pole at

s=-2

moved left to

s=-10

when the loop closed (faster response), and the dc gain became

T(0)=8/10=0.8

— both consistent with high-gain negative feedback.

1 + \dfrac{8}{s+2} = 0 \;\Rightarrow\; s = -10\ (\text{stable})

Common pitfalls

•Testing the open-loop denominator for stability. You must form the closed-loop characteristic equation $1+GH=0$ and clear fractions first; the open-loop poles are not the closed-loop poles.
•Treating a pole on the imaginary axis as stable. $\text{Re}(p)=0$ is marginally stable (sustained oscillation), not BIBO stable, which requires strictly $\text{Re}(p)<0$ .
•Concluding stability from all-positive coefficients. That is necessary but not sufficient for order $\ge 3$ ; the fourth-order example passes the coefficient test yet has two RHP roots.
•Sign-error in the Routh determinant. The numerator is (pivot row times the column term above-right) minus (top entry times the next term) — i.e. the negative of the literal 2x2 determinant; divide by the pivot, and keep the cross-multiplication order consistent.
•Conflating the linear and dB gain margins. The bare factor $K_{\text{marg}}/K_{\text{design}}$ is the linear gain margin; the dB margin is $20\log_{10}$ of that ratio (a factor of 5 is about 14 dB, not 5 dB). Never equate the unitless ratio to the $-20\log_{10}|G(j\omega_{180})|$ value.
•Reading the root locus backward: branches begin at open-loop poles and terminate at open-loop zeros (or infinity) as $K$ increases, not the reverse.

References

NCEES FE Reference Handbook — Instrumentation, Measurement, and Control: Control Systems — Closed-loop characteristic equation $1+G_1G_2H=0$ , Final Value Theorem dc gain, and gain/phase-margin definitions.
NCEES FE Reference Handbook — Electrical and Computer Engineering: Decibels and Bode Plots — Frequency-response basis for the gain-margin computation.
Ogata, Modern Control Engineering — Routh-Hurwitz array construction and root-locus rules (off-handbook supplements for area C), ch. 5-6.

D. Controller performance

Controller Performance and PID

Compute steady-state error from system type and error constants, read transient specs off the damping ratio, and tune P/PI/PID actions.

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